Theorem nss 4003

Description: Negation of subclass relationship. Exercise 13 of [TakeutiZaring] p. 18. (Contributed by NM, 25-Feb-1996.) (Proof shortened by Andrew Salmon, 21-Jun-2011.)

Assertion

Ref	Expression
nss	⊢ (¬ 𝐴 ⊆ 𝐵 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Proof of Theorem nss

Step	Hyp	Ref	Expression
1		exanali 1882	. . 3 ⊢ (∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) ↔ ¬ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
2		df-ss 3924	. . 3 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
3	1, 2	xchbinxr 338	. 2 ⊢ (∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) ↔ ¬ 𝐴 ⊆ 𝐵)
4	3	bicomi 227	1 ⊢ (¬ 𝐴 ⊆ 𝐵 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 400  ∀wal 1561  ∃wex 1802   ∈ wcel 2145   ⊆ wss 3907

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1818  ax-4 1832

This theorem depends on definitions:  df-bi 210  df-an 401  df-ex 1803  df-ss 3924

This theorem is referenced by:  nssrex  4004  grur1  10793  psslinpr  11004  reclem2pr  11021  mreexexlem2d  17691  prmcyg  19955  filconn  24001  alexsubALTlem4  24168  wilthlem2  27191  shne0i  31709  onvf1odlem2  35459  erdszelem10  35563  fundmpss  36130  ntrneineine1lem  44672  nssd  45681  nsssmfmbf  47351